Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 386: 55


$$ y'= (4x-8)e^{(x-1)^2}e^{(x-3)^2}.$$

Work Step by Step

Taking the $\ln $ on both sides of the equation, we get $$\ln y= \ln \left(e^{(x-1)^2}e^{(x-3)^2}\right)$$ Then using the properties of $\ln $, we can write $$\ln y= \ln e^{(x-1)^2}+\ln e^{(x-3)^2}\\ =(x-1)^2+(x-3)^2.$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= 2(x-1)+2(x-3)=4x-8,$$ Hence $ y'$ is given by $$ y'=y(4x-8)=(4x-8)e^{(x-1)^2}e^{(x-3)^2}$$
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